Why don't parabolas have asymptotes?

an asymptote is a line which becomes the tangent of the curve as the $$x$$ or $$y$$ cordinates of the curve tends to infinity.

Hyperbola has asymptotes but parabolas ( both being an open curve and a conic section) do not.

I am just curious to know why don't parabolas have an asymptote ? Is there any mathematical proof to show that ?

• How do you define a parabola? Sep 5 '20 at 5:24
• @Andrew Chin it is the locus of points whose distance from a fixed line and from a fixed point is equal. Sep 5 '20 at 5:40
• Add a link to the page Sep 5 '20 at 5:56
• Hyperbolas are the only conic sections with asymptotes. Even though parabolas and hyperbolas look very similar, parabolas are formed by the distance from a point and the distance to a line being the same. Therefore, parabolas don’t have asymptotes. Sep 5 '20 at 6:58
• @SarGe this is the line which I couldn't understand. Can you make it clear ? Sep 5 '20 at 7:00

If a parabola had an asymptote, then we could translate-rotate it so that the asymptote coincides with the $$y$$ axis.

Now, consider the general conic section (which, of course, includes all rotated parabolas):

$$A x^2 + B x y + C y^2 + D x + E y + F= 0 \tag1$$ Or $$y( C y + B x + E) + A x^2 + D x + F=0 \tag2$$

and let's see under what conditions it could happen its graphic has the $$y$$ axis as an asymptote, i.e. that as $$x\to 0$$ $$y \to \pm\infty$$.

In that limit, we must have $$( C y + B x + E) \to -\frac{F}{y} \to 0\tag3$$

But this requires $$C=0$$ and $$E=0$$

Then it cannot be a parabola (or an ellipse), only an hyperbola.

• Great answer! But for $( C y + B x + E) \to 0$, isn't $y \to \frac{-E}{C}$ sufficient? And for that, the only conclusions I can see are $E \to \pm \infty$, $C \to 0$ or both. How is it only $C=E=0$ that you got as a solution? Admittedly, I've never heard anyone else talk about coefficients being limits, but this is what I can conclude analytically. Where did I go wrong? Aug 26 '21 at 15:40
• @harry When want both $x\to 0$ and $y \to \infty$ to happen. Aug 26 '21 at 17:20
• But isn't that the condition that's required? That's what you've used in your answer, right? Aug 26 '21 at 18:17

Consider a hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ Any point on it is $$(a\sec\theta,b\tan\theta)$$. Now, asymptote is tangent at point at infinity i.e. $$\theta\to \pm π/2$$.

A general tangent to a hyperbola is $$\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1\\ \implies \frac{x}{a}-\frac{y\sin\theta}{b}=\cos\theta$$ If $$\theta\to \pm π/2$$, we have $$y=\pm\frac{b}{a}x$$

Consider a parabola $$y^2=4ax$$ whose parametric point is $$(at^2,2at)$$. Its general tangent is $$y=\frac{x}{t}+at.$$ Now, if $$t\to\infty$$, we have equation of tangent as $$y=\infty$$.

In analytic geometry, an asymptote of a curve is a line such that the distance between the curve and the line approaches zero as one or both of the $$x$$ or $$y$$ coordinates tends to infinity.

But in this case, the distance is already infinity and doesn't approach it. Hence, there is no asymptote for parabola. Similarly, there are no asymptotes for ellipse, too.

Hints:

An $$\textit{asymptote}$$ is a straight line to which the curve approaches while it moves away from the origin. The curve can approach the line from one side, or it can intersect it again and again. Not every curve which goes infinitely far from the origin (infinite branch of the curve) has an asymptote.

The graph of the function $$y=ax^n$$

where $$\space n>0, \space$$ integer, is a $$\textit{parabola of n-th degree, or of n-th order}.$$

For functions given in explicit form $$\space y=f(x), \space$$ we know: the vertical asymptotes are at points of discontinuity where the function $$\space f(x)\space$$ has an infinite jump; the horizontal and oblique asymptotes have the equation:

$$y=kx+b, \space \text{ with } \space \space k=\lim_{x \to \infty}\frac{f(x)}{x}, \space \space b=\lim_{x \to \infty}\left[f(x)-kx \right].$$